Let BC be the height of the tower and A be the position of the ball lying on the ground. Then,
BC = 20 m and \(\angle\)XCA = 60° = \(\angle\)CAB
Let AB = x metres.
In the right angled ΔABC,
tan 60° = \(\frac { BC }{ AB } \)
\(\sqrt { 3 } =\frac { 20 }{ x } \)
\(x=\frac { 20\times \sqrt { 3 } }{ \sqrt { 3 } \times \sqrt { 3 } } =\frac { 20\times 1.732 }{ 3 } \) = 11.54m
Hence, the distance between the foot of the tower and the ball is 11.55 m.

Let AD is the first building.
BC is the second building.
AD = BE = BC - CE
o
From the right triangle CED
\(\tan 45^{\circ} =\frac{C E}{D E} \)
\(1 =\frac{C E}{A B}=\frac{C E}{70 m} \)
CE = 70 m
BE = BC - EC
= 120 m - 70 m = 50 m
AD = 50 m
Height of the first building is 50 m.

Let AB be the rock.
C be the position of the car.
In right triangle ABC
\(\tan 30^{\circ} =\frac{A B}{B C} \)
\(\frac{1}{\sqrt{3}} =\frac{50 \sqrt{3}}{B C} \)
\(B C =50 \sqrt{3} \times \sqrt{3} \)
= 50 x 3 = 150 m
Distance of the car from the rock = 150 m.

Let AB = x be the distance between foot of the house and the observer at the median of the road.
DB = 2x is the width of the road.
Height of the house BC = \(4 \sqrt{3} m\)
From the right triangle \(\triangle\) ABC
\( \therefore \tan 30^{\circ} =\frac{B C}{A B} \)
\(\frac{1}{\sqrt{3}}=\frac{4 \sqrt{3}}{x} \)
\(x =4 \sqrt{3} \times \sqrt{3}=4 \times 3=12 \mathrm{~m} \)
\(\text { Width of the road } =2 \times x=2 \times 12=24 \mathrm{~m}\)
Width of the road = 24 m.

From the right \(\triangle\)ABC
\( \tan \theta =\frac{\text { Opposite side }}{\text { Adjacent side }}=\frac{A C}{B C} \)
\(=\frac{10 \sqrt{3} m}{30 m}=\frac{\sqrt{3}}{3} \)
\(=\frac{\sqrt{3}}{\sqrt{3} \sqrt{3}}=\frac{1}{\sqrt{3}} \)
\(\tan \theta =\frac{1}{\sqrt{3}} \)
\(\theta =\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
0 = 30^o
Angle of elevation is 30^o

Let PQ the height of the tower.
Take PQ = h and QR is the distance between the tower and the point R.in right triangle PQR,\(\angle \)PRQ=30°
tan\(\theta =\frac { PQ }{ QR } \)
tan30° = \(\frac { h }{ 48 } \) gives,\(\frac { 1 }{ \sqrt { 3 } } =\frac { h }{ 48 } \) so, h =\(16\sqrt { 3 } \)
Therefore the height of the tower \(16\sqrt { 3 } \) m

in the right triangle ABC [see figure. (a)]
tan \(\theta \) =\(\frac { opposite\ side\ }{ adjacent\ side\ } =\frac { 4 }{ 5 } \)
= tan^-1(0.8)
\(\theta \) = \(38.7°\)(since tan \(38.7°\) = 0.8011)
\(\angle \)BAC = \(38.7°\)

\(\frac { sec\theta }{ sin\theta } -\frac { sin\theta }{ cos\theta } = \frac { \frac { 1 }{ cos\theta } }{ sin\theta } -\frac { sin\theta }{ cos\theta } =\frac { 1 }{ sin\theta cos\theta } -\frac { sin\theta }{ cos\theta } \)
\(=\frac { 1-si{ n }^{ 2 }\theta }{ sin\theta cos\theta } =cot\theta \)

\(\sqrt { \frac { 1+cos\theta }{ 1-cos\theta } } \)=\(\sqrt { \frac { 1+cos\theta }{ 1-cos\theta } \times \frac { 1+cos\theta }{ 1+cos\theta } } \) [multiply numerator and denominator by the conjugate of 1 - cos\(\theta \)]
=\(\sqrt { \frac { (1+cos\theta { ) }^{ 2 } }{ (1-cos\theta { ) }^{ 2 } } } \) =\(\frac { 1+cos\theta }{ \sqrt { si{ n }^{ 2 }\theta } } \) [since sin²\(\theta \) + cos²\(\theta \) = 1]
=\(\frac { 1+cos\theta }{ sin\theta } =cosec\theta +cot\theta \)

tan² \(\theta \) - sin ²\(\theta \) = tan² \(\theta \) -\(\frac { si{ n }^{ 2 }\theta }{ co{ s }^{ 2 }\theta } \),cos²\(\theta \) 
= tan² \(\theta \) (1-cos² \(\theta \) ) = tan² \(\theta \) sin^{2 \(\theta \)}